1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Second derivative=0; stable/unstable equilibrium?

  1. Oct 4, 2009 #1
    1. The problem statement, all variables and given/known data

    Describe how to determine whether an equilibrium is stable or unstable when [d2U/dx2]_0 = 0

    From Classical Dynamics - Ch 2 #45 - Marion Thornton

    2. Relevant equations AND 3. The attempt at a solution

    When second derivative positive, equilibrium is stable. When second derivative negative, equilibrium unstable.

    When second derivative is 0, you have to do another test. I was under the impression that you could find a third derivative, fourth derivative, until you get a positive/negative value for x=0 and use the results to get stability, with the same rule as above? And if second and subsequent derivatives are 0, it's a neutral equilibrium.

    The main reason I'm confused is the book had a weird solution that didn't make sense to me. They expanded the potential about the equilibrium point (how? why?), found the first derivative, and said that it's stable when n is even and unstable when n is odd. But that doesn't make sense to me. How could they make a conclusion like that if they don't know what U is?

    Maybe I'm misunderstanding the question's notation - I understood it to mean "second derivative at x=0 is 0."
     
  2. jcsd
  3. Oct 4, 2009 #2

    kuruman

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    What is n? Is your potential of the form U = kxn (n > 0)? If so, sketch the potential for even and odd values of n and see for yourself what happens at x = 0.
     
  4. Oct 5, 2009 #3

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    Ermmm...what the heck is a "neutral equilibrium"?:confused:

    I'm guessing that the expansion is a Taylor expansion...you know how to do those right? What is the derivative of potential equal to physically? What should that quantity be near a stable/unstable equilibrium point?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Second derivative=0; stable/unstable equilibrium?
  1. Unstable Equilibrium (Replies: 21)

Loading...